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In order to properly state the question let me be precise about the object at the core of this question's title. First, given any orthonormal basis of $G:=\{G_j\}_{j=1}^{n^2}$ of $\mathbb C^{n\times n}$ (equipped with the Hilbert-Schmidt inner product) the corresponding matrix representation of a linear map $\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$ with respect to $G$ is given by $$\sum_{j,k=1}^{n^2}{\rm tr}(G_j^\dagger\Phi(G_k))|j\rangle\langle k|\in\mathbb C^{n^2\times n^2}\,.$$ Important special cases for this are the natural representation $K(\Phi)$ or $\widehat\Phi$—which comes from choosing $G$ to be the computational basis $\{|j\rangle\langle k|\}_{j,k=1}^n$—or, in case $n=2^d$ for some $d$, the Pauli transfer matrix $\mathcal P(\Phi)$—for which one chooses $G$ to be all elementary tensor products of single Paulis $\frac1{\sqrt2}\sigma_j$, $j=0,\ldots,3$ (where $\sigma_0={\bf1}$).

Of course this concept makes sense for linear maps between matrices of any size but for the sake of this question we restrict ourselves to the simplest case of one qubit, i.e. $n=2$. In this case it is easy to see that if $\Phi$ is Hermitian preserving (i.e. $\Phi(X^\dagger)=\Phi(X)^\dagger$ for all $X$) and trace preserving, then there exists $v\in\mathbb R^3$ and $\Lambda\in\mathbb R^{3\times 3}$ such that $$ \mathcal P(\Phi)=\begin{pmatrix}1&0\\v&\Lambda\end{pmatrix}\,. $$ While, at this stage, the question of uniqueness does not really make sense yet it will once we ge to the so-called normal form. For this observe that the Pauli transfer matrix of a unitary channel $X\mapsto UXU^\dagger$ is of the block-diagonal form ${\rm diag}(1,R)$ with $R\in\mathsf{SO}(3)$. In fact, this gives rise to an isomorphism between the group ${\rm Ad}_{\mathsf{SU}(2)}$ and $\mathsf{SO}(3)$ (cf. Appendix B in J. Schlienz and G. Mahler, Description of Entanglement, Phys. Rev. A, 52 (1995)), that is, for every $R\in\mathsf{SO}(3)$ there even exists (basically unique) $U\in\mathsf{SU}(2)$ such that $\mathcal P(U(\cdot)U^\dagger)={\rm diag}(1,R)$. This gives rise to the

Normal form. For every linear, Hermitian-preserving, and trace-preserving map $\Phi:\mathbb C^{2\times 2}\to\mathbb C^{2\times 2}$ there exist $U,V\in\mathsf{SU}(2)$ and $v,\lambda\in\mathbb R^3$ with $v_1,v_2\geq 0$ and $\lambda_1\geq\lambda_2\geq|\lambda_3|$ such that $$ \mathcal P(V^\dagger\Phi(U(\cdot)U^\dagger)V)=\begin{pmatrix}1&0&0&0\\v_1&\lambda_1&0&0\\v_2&0&\lambda_2&0\\v_3&0&0&\lambda_3\end{pmatrix}\,,\tag{1} $$

cf., e.g., F. Verstraete and H. Verschelde, On Quantum Channels, 2002 and M. Wolf and J. Cirac, Dividing Quantum Channels, Commun. Math. Phys., 279 (2008) (arXiv). This raises the following

Question: Is this normal form unique, that is, given some $\Phi$ as well as two pairs $(v^1,\lambda^1), (v^2,\lambda^2)$ which both satisfy the previously stated constraints and which both constitute a normal form (1) of $\Phi$ does this imply $v^1=v^2$, $\lambda^1=\lambda^2$?

What is easy to see is that $\lambda^1=\lambda^2$ as they are (basically) the singular values of $\Lambda$. However, it is not immediately clear whether the $v^j$ admit an additional degree of freedom. While the previously quoted paper of Verstraete & Verschelde claims uniqueness (p.9 therein) they do not give a proof which is what motivated this question.


(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)

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It turns out that uniqueness can only be guaranteed if $\lambda_1>\lambda_2>|\lambda_3|$; otherwise one can construct counterexamples (as we will do below).

In order to understand why uniqueness may fail we have to recall how the normal form is established in the first place. The main idea is to singular value decompose $\Lambda$ (however, $\lambda_3$ may be negative as one "only" has access to $\mathsf{SO}(3)$ instead of $\mathsf O(3)$ so it it not a "proper" SVD). Now for the standard singular value decomposition $A=V\Sigma W^*$ of a square matrix $A$ it holds that

  • $V$ is unique (up to a diagonal matrix containing phase factors) if the singular values of $A$ differ pairwise,
  • if $V$ is fixed and if $A$ is full rank, then $W$ is uniquely determined,

refer, e.g., to Theorem 7.3.5 in Horn & Johnson's book "Matrix Analysis" (first edition), 1987. The first point is what's important for us: if $\lambda_1>\lambda_2>|\lambda_3|$, then the $R_1\in\mathsf{SO}(3)$ applied to $\mathcal P(\Phi)$ from the left is unique (up to a diagonal matrix). Moreover, $R_1$ is what transforms $v$ so because we demanded $v_1,v_2\geq 0$ we can conclude $v^1=v^2$ in this case (because $\det R_1=1$ two signs would have to be changed simultaneously).

This analysis also makes it clear when uniqueness fails: some of the $|\lambda_j|$ have to coincide and at least one of the corresponding $v_j$ has to be non-zero. With this we come to the promised counterexample: Consider $\Phi$ such that $$ \mathcal P(\Phi)=\begin{pmatrix} 1&0&0&0\\v_1&\lambda_1&0&0\\v_2&0&\lambda_2&0\\v_3&0&0&-\lambda_2 \end{pmatrix} $$ for some $v\in\mathbb R^3$, $\lambda_1>\lambda_2\geq 0$ such that $v_2\neq 0$. We claim that there then exist $U,V\in\mathsf{SU}(2)$ such that $$ \mathcal P(V^\dagger\Phi(U(\cdot)U^\dagger)V)=\begin{pmatrix} 1&0&0&0\\v_1&\lambda_1&0&0\\0&0&\lambda_2&0\\\sqrt{v_2^2+v_3^2}\;{\rm sgn}(v_3)&0&0&-\lambda_2 \end{pmatrix}\tag{2} $$ meaning we would have found two normal forms of $\Phi$ which clearly differ from each other. Indeed, if we choose $U,V$ such that $$ \mathcal P(U(\cdot)U^\dagger)=\begin{pmatrix} 1&0&0\\0&1&0\\0&0&R_{\arctan(-v_2/v_3)} \end{pmatrix}\qquad\mathcal P(V(\cdot)V^\dagger)= \begin{pmatrix} 1&0&0\\0&1&0\\0&0&R_{\arctan(v_2/v_3)} \end{pmatrix} $$ where $$ R_\phi:=\begin{pmatrix} \cos\phi&\sin\phi\\-\sin\phi&\cos\phi \end{pmatrix}\in\mathsf{SO}(2) $$ for all $\phi\in\mathbb R$ is the usual rotation matrix, then (2) holds. The reason for this is that $R_\phi\sigma_z R_\phi=\sigma_z$ for all $\phi$ (so the $\Lambda$ block is left invariant) while $\mathcal P(U(\cdot)U^\dagger)$ acts on $v$ via \begin{align*} R_{\arctan(-v_2/v_3)}\begin{pmatrix} v_2\\v_3 \end{pmatrix}&=\frac{1}{v_3\sqrt{1+\frac{v_2^2}{v_3^2}}}\begin{pmatrix} v_3 & -v_2 \\ v_2 & v_3 \end{pmatrix}\begin{pmatrix} v_2\\v_3 \end{pmatrix}\\ &=\begin{pmatrix} 0\\\sqrt{1+\frac{v_2^2}{v_3^2}}v_3 \end{pmatrix}=\begin{pmatrix} 0\\\sqrt{v_2^2+v_3^2}\;{\rm sgn}(v_3) \end{pmatrix}\,. \end{align*}

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